00:01
So you have a wild type amino acid sequence that i copied from the board already.
00:07
You have seven different amino acids.
00:10
There are methanine, glycine, glutamine, threatening, lysine, valiant.
00:16
So we know that the mutant form of a short protein from e.
00:20
Coli will be generated.
00:22
A change in the dna sequence of the gene has occurred in each of these mutations.
00:27
You have three different mutant.
00:30
One has methanine only.
00:32
Mutant 2 has methanine, glycine, glutamine, and the threading is changed in asperogen.
00:41
And mutant number 3 has all the sequence until the valine.
00:45
Now, valine become aline.
00:47
So the question says that it could be either a single -base pair change or a single -base -pair insertion or deletion.
00:56
Determining the base sequence change in this wild -type dna template strength that could have lead to each of the mutant 1, 2, and 3.
01:06
So as you can see, that methadine always is going to be in dna atg.
01:14
So glycine has four different possible codons.
01:20
So let's say you have g -g -g -u, g -g -t -m actually, and g -g -g -a, g -g -g -g -e.
01:40
C and g g g they all give you the same amino acid which is glycine.
01:52
So if we look at mutant 1 so mutant 1 has only a methionine which means that the glycine in mutant 1 change into a stop codon.
02:07
So if we look at all the stop codon there are three possible stop codon.
02:13
So let's say mutant 1.
02:19
So stop codon can be uga, uag, and uaaaa.
02:29
So if we look at all the codounds of glycine and then you'll look at the stop codon, most likely the original sequence is going to be a gga, which give you a glycine.
02:46
However, due to a single point mutation, a g is becoming a u or in this case in this is remember this is m rna but in dna it's going to be a t t g a g a g or taa this is going to be the dna sequence so in this case we are we can deduce that most likely the force nucleotide g a g a g a turn into a tg.
03:34
So if we look at the dna sequence 5, atg will give you methanine in the protein, and the next one is gonna be a tga instead of gga.
03:53
So this will give you a stock codon, which is uga in m rna, so that the protein actually terminated right there so that you only have one amino acid which is methionine so now we have already figured out the mutant number one so which mostly like might likely that a g is turned into a t in mutant one all right so now let's move on to question on mutant two so mutant two has methine glycine glycine but theory become a spartic asparagid and then the rest of the sequence is gone so the four possible threatening codon will be acu acc aca and acg so and again in dna it's gonna be a t and if we also take a look at the lysine.
05:29
There are two possible codon for lysine.
05:31
One is aaa, the other is a .a.
05:47
So in this case, most likely there is an insertion mutation because threatening is now become asparagus.
06:01
So my deduction is that the codon for threatening must be a a act in the dna and it's going to be a aaa or an enlycine.
06:16
So let's say originally it has a act and taa.
06:22
Now if you have a insertion between a and c, let's say an insertion here is a a.
06:33
So let's say mutant 2.
06:37
So the rest of the sequence are exactly the same, but when it get to threatening and lysine.
06:47
Now, originally, again, it is a act, and this is a, a, let's say.
06:59
Now, if you have an insertion right here with an a, so the new sequence is going to be a, a, c, and t -a.
07:13
So this caused a -c -t, codon become a a -a -c.
07:19
The a -a -c will actually give you the asperogen asprag, and this will also cause a frame -shift mutation.
07:32
So a -a, licensing is gone.
07:35
Instead, this t here is pushing towards the next codon.
07:40
So you have a new codon.
07:42
Taa, taa is actually uaa in m -r -n -a, which will be a stop.
07:47
And this is why the rest of the sequence will not be translated, a stop at the asperogen.
07:54
So most likely there is a insertion there in between the original threading, hold on, act, so this insertion caused the frame shift mutation, so that you can see that the polypeptide will actually terminate earlier due to introduction of the stop codon right here.
08:28
So this is, again, why the mutant two has what it looks like.
08:33
So when we look at the number of nucleotide, so you have 3, 3, 3, and then number 10, after the number 10 nucleotide, you have a single, nucleotide insertion of nucleotide a.
09:02
So this caused the frame -shipped mutation that, which happens in mutant 2...